Finding the graph formula with know points and other equation.

  1. Mar 11, 2014 #1
    I have an equation for determining the acceleration of an object being propelled by a constant power source, that is affected by air resistance:

    a = [itex]\frac{P}{mv}[/itex]-[itex]\frac{CDpAv2}{2m}[/itex]

    Since F = [itex]\frac{P}{v}[/itex]

    I am trying to graph this as a velocity-time graph, however, I don't know how to do it. There is no time variable that I can replace with x, and the y-value (velocity) is mixed into the equation already.

    I remember the equation given to me for a similar sort of thing, without air resitance, but instead just a constant friction force, that was something like this:

    y = k(1-e-ax)

    Where k is a constant, which is the maximum speed, and a is another constant which represents the force of air resistance.

    The maximum speed in this case is [itex]\sqrt[3]{\frac{2P}{CDpA}}[/itex], so the equation would be something like:

    y = [itex]\sqrt[3]{\frac{2P}{CDpA}}[/itex](1-e-ax)

    But that's as far as I've gotten. By the use of iteration, I have determined the velocity at several different times, here's a few, just in case it helps:

    (0,0) (25,83.4762) (50,118.1195) (75,126.1601) (100,127.7624) (125,128.0709)

    This is for variables of values: p = 0.001, A = 1900.933, m = 1900.933, P = 1*106
    and CD = 0.5

    Any help would be greatly appreciated, and if you need any more information, just let me know.

    EDIT: the equations don't seem to be formatting correctly, so I'll redo them down here:

    a = P/v - (Cd*p*A*v^2) /2
    F = P/v
    max speed = ( (2*P) / (Cd*p*A) )^1/3
    y = ( (2*P) / (Cd*p*A) )^1/3 * (1 - e^-ax)
    Last edited: Mar 11, 2014
  2. jcsd
  3. Mar 11, 2014 #2
    Given ##a=\frac{dv}{dt}##, what you have is a separable ordinary differential equation [tex]\frac{dv}{dt}=\frac{\alpha}{v}-\beta v^2[/tex] where ##\alpha=\frac{P}{m}## and ##\beta=\frac{C_DpA}{2m}## have been substituted to make the equation (slightly) more manageable. Hopefully that's enough to get you going.

    I'm not sure where you got your force equation. Are you not using ##F=ma##? Furthermore, I don't know that it's even useful here.

    I reckon it's doubtful that the equation that works in a constant friction situation applies here. So you're likely barking up the wrong tree there.
  4. Mar 12, 2014 #3
    I have never done differential equations before, and I tried to just find the anti-derivative, but my teacher said that I had done it incorrectly, as I differentiated dv instead of dt, or the other way around. Anyway, here is what I got:

    v = (Pt/m)*ln(|v|)-(CDpAv3t)/(6m)+c

    Unfortunately from there, if it is correct, I don't know where to go - how to get v by itself without v on the other side.

    What do I do from here, or from the start if that isn't correct?

    Thanks for your response.
    Last edited: Mar 12, 2014
  5. Mar 12, 2014 #4

    Stephen Tashi

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    Science Advisor

    I think this is due to your using the SUB and SUP tags instead of the usual LaTex notation.

    [itex] a = \frac{P}{v} - \frac{ (C_d p A v^2) }{2} [/itex]
  6. Mar 12, 2014 #5
    Oh ok, I'll give it a go:

    a = ([itex]\frac{Pt}{m}[/itex])ln(|v|)-[itex]\frac{C_{D}pAv^{3}t}{6m}[/itex]

    Seems to work, thanks.
  7. Mar 16, 2014 #6
    Does anyone have any advice on what I should do to get time into the equation, without velocity on both sides? I don't even know what to type into google to find information on it. Any information would be much appreciated.
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